If the HCF of two numbers is 6 and their LCM is 120, one such pair of numbers is:

Options

A. 24, 30
B. 18, 40
C. `12, 60`
D. `12, 40`

Answer

To solve the problem, we need to find two numbers given that their Highest Common Factor (HCF) is 6 and their Least Common Multiple (LCM) is 120.

### Step-by-Step Solution:

1. **Understand the Relationship Between HCF and LCM**:
The relationship between the HCF (also known as GCD) and LCM of two numbers \( a \) and \( b \) is given by:
\[ \text{HCF} \times \text{LCM} = a \times b \]
In this case, we know:
- HCF = 6
- LCM = 120

2. **Set Up the Equation**:
Using the relationship, we can set up the equation:
\[ 6 \times 120 = a \times b \]
Simplifying this gives:
\[ 720 = a \times b \]

3. **Express the Numbers in Terms of HCF**:
Since the HCF is 6, we can express the two numbers as:
\[ a = 6x \quad \text{and} \quad b = 6y \]
where \( x \) and \( y \) are coprime integers (they have no common factors other than 1).

4. **Substituting into the Product Equation**:
Substitute \( a \) and \( b \) into the product equation:
\[ (6x) \times (6y) = 720 \]
This simplifies to:
\[ 36xy = 720 \]
Dividing both sides by 36 gives:
\[ xy = 20 \]

5. **Finding Coprime Pairs**:
Now we need to find pairs of integers \( (x, y) \) such that their product is 20 and they are coprime. The pairs of factors of 20 are:
- (1, 20)
- (2, 10)
- (4, 5)

The coprime pairs from the above are:
- (4, 5)

6. **Calculating the Original Numbers**:
Using the coprime pair \( (4, 5) \):
\[ a = 6x = 6 \times 4 = 24 \]
\[ b = 6y = 6 \times 5 = 30 \]

7. **Final Result**:
Therefore, one such pair of numbers is:
\[ (24, 30) \]

### Final Answer:
The pair of numbers is \( 24 \) and \( 30 \).